Suppose that $f$ is a function such that $f(a) \leq f(b)$ whenever $a < b$.
Prove that $\displaystyle\lim_{x\to a^{-}}f(x)$ and $\displaystyle \lim_{x\to a^{+}}f(x)$ both exists.

Here is my work so far for $\displaystyle\lim_{x\to a^{-}}f(x)$:

Considering the interval $(-\infty, a)$. Since $f(a) \leq f(b)$ whenever $a < b$ it is clear that for every $x\in (-\infty, a)$ $f(x) \leq f(a)$. That is $f(x)$ is bounded above at $(-\infty, a)$ by $f(a)$. Since $f(x)$ is bounded above, $f(x)$ has a least upper bound, let it be $\alpha$.